In an isothermal process, the pressure and volume follow Boyle's Law: \[ P V = \text{constant} \] whereas in an adiabatic process, the relation follows Poisson's equation: \[ P V^\gamma = \text{constant} \] For a given ratio of heat capacities \( \gamma = 1.5 \), using these relations, we can derive the pressure ratio as \( 2:1 \). Understanding these fundamental gas laws helps in solving thermodynamics problems efficiently.
Step 1: Understanding the Given Conditions - For an isothermal process, the pressure-volume relation is given by Boyle’s Law: \[ P V = \text{constant} \] Thus, if volume is doubled, the new pressure is: \[ P_2 = \frac{P_{\text{initial}}}{2} \] - For an adiabatic process, the pressure-volume relation is given by Poisson’s equation: \[ P V^\gamma = \text{constant} \] where \( \gamma \) is the adiabatic index, given as 1.5.
Step 2: Applying the Adiabatic Equation For an adiabatic process: \[ P_{\text{initial}} V^\gamma = P_1 (2V)^\gamma \] Rearranging: \[ P_1 = P_{\text{initial}} \times (2)^{-\gamma} \] Substituting \( \gamma = 1.5 \): \[ P_1 = P_{\text{initial}} \times (2)^{-1.5} = \frac{P_{\text{initial}}}{2^{1.5}} \] Since \( 2^{1.5} = \sqrt{8} = 2\sqrt{2} \), we get: \[ P_1 = \frac{P_{\text{initial}}}{2\sqrt{2}} \] \
Step 3: Equating \( P_1 \) and \( P_2 \) Given that \( P_1 = P_2 \), we equate the expressions: \[ \frac{P_{\text{initial}}}{2\sqrt{2}} = \frac{P_{\text{initial}}}{2} \] Solving for the pressure ratio: \[ \frac{P_{\text{initial}}}{P_{\text{initial (iso)}}} = \sqrt{2}:1 \] Thus, the correct ratio is: \[ \mathbf{\sqrt{2}:1} \]
A cube of side 10 cm is suspended from one end of a fine string of length 27 cm, and a mass of 200 grams is connected to the other end of the string. When the cube is half immersed in water, the system remains in balance. Find the density of the cube.

